3.281 \(\int \frac{x^m (a+b \sin ^{-1}(c x))^2}{(d-c^2 d x^2)^3} \, dx\)

Optimal. Leaf size=668 \[ -\frac{b^2 c^2 (1-m) (m+1) x^{m+3} \text{HypergeometricPFQ}\left (\left \{1,\frac{m}{2}+\frac{3}{2},\frac{m}{2}+\frac{3}{2}\right \},\left \{\frac{m}{2}+2,\frac{m}{2}+\frac{5}{2}\right \},c^2 x^2\right )}{6 d^3 \left (m^2+5 m+6\right )}-\frac{b^2 c^2 (3-m) (m+1) x^{m+3} \text{HypergeometricPFQ}\left (\left \{1,\frac{m}{2}+\frac{3}{2},\frac{m}{2}+\frac{3}{2}\right \},\left \{\frac{m}{2}+2,\frac{m}{2}+\frac{5}{2}\right \},c^2 x^2\right )}{4 d^3 \left (m^2+5 m+6\right )}+\frac{b c (1-m) (m+1) x^{m+2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 (m+2)}+\frac{b c (3-m) (m+1) x^{m+2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 (m+2)}+\frac{(1-m) (3-m) \text{Unintegrable}\left (\frac{x^m \left (a+b \sin ^{-1}(c x)\right )^2}{d-c^2 d x^2},x\right )}{8 d^2}+\frac{b^2 c^2 (1-m) x^{m+3} \text{Hypergeometric2F1}\left (1,\frac{m+3}{2},\frac{m+5}{2},c^2 x^2\right )}{6 d^3 (m+3)}+\frac{b^2 c^2 (3-m) x^{m+3} \text{Hypergeometric2F1}\left (1,\frac{m+3}{2},\frac{m+5}{2},c^2 x^2\right )}{4 d^3 (m+3)}+\frac{b^2 c^2 x^{m+3} \text{Hypergeometric2F1}\left (2,\frac{m+3}{2},\frac{m+5}{2},c^2 x^2\right )}{6 d^3 (m+3)}+\frac{(3-m) x^{m+1} \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}+\frac{x^{m+1} \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac{b c (1-m) x^{m+2} \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \sqrt{1-c^2 x^2}}-\frac{b c (3-m) x^{m+2} \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \sqrt{1-c^2 x^2}}-\frac{b c x^{m+2} \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \left (1-c^2 x^2\right )^{3/2}} \]

[Out]

-(b*c*x^(2 + m)*(a + b*ArcSin[c*x]))/(6*d^3*(1 - c^2*x^2)^(3/2)) - (b*c*(1 - m)*x^(2 + m)*(a + b*ArcSin[c*x]))
/(6*d^3*Sqrt[1 - c^2*x^2]) - (b*c*(3 - m)*x^(2 + m)*(a + b*ArcSin[c*x]))/(4*d^3*Sqrt[1 - c^2*x^2]) + (x^(1 + m
)*(a + b*ArcSin[c*x])^2)/(4*d^3*(1 - c^2*x^2)^2) + ((3 - m)*x^(1 + m)*(a + b*ArcSin[c*x])^2)/(8*d^3*(1 - c^2*x
^2)) + (b*c*(1 - m)*(1 + m)*x^(2 + m)*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, c^2*x^2
])/(6*d^3*(2 + m)) + (b*c*(3 - m)*(1 + m)*x^(2 + m)*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, (2 + m)/2, (4 +
m)/2, c^2*x^2])/(4*d^3*(2 + m)) + (b^2*c^2*(1 - m)*x^(3 + m)*Hypergeometric2F1[1, (3 + m)/2, (5 + m)/2, c^2*x
^2])/(6*d^3*(3 + m)) + (b^2*c^2*(3 - m)*x^(3 + m)*Hypergeometric2F1[1, (3 + m)/2, (5 + m)/2, c^2*x^2])/(4*d^3*
(3 + m)) + (b^2*c^2*x^(3 + m)*Hypergeometric2F1[2, (3 + m)/2, (5 + m)/2, c^2*x^2])/(6*d^3*(3 + m)) - (b^2*c^2*
(1 - m)*(1 + m)*x^(3 + m)*HypergeometricPFQ[{1, 3/2 + m/2, 3/2 + m/2}, {2 + m/2, 5/2 + m/2}, c^2*x^2])/(6*d^3*
(6 + 5*m + m^2)) - (b^2*c^2*(3 - m)*(1 + m)*x^(3 + m)*HypergeometricPFQ[{1, 3/2 + m/2, 3/2 + m/2}, {2 + m/2, 5
/2 + m/2}, c^2*x^2])/(4*d^3*(6 + 5*m + m^2)) + ((1 - m)*(3 - m)*Unintegrable[(x^m*(a + b*ArcSin[c*x])^2)/(d -
c^2*d*x^2), x])/(8*d^2)

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Rubi [A]  time = 0.91303, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{x^m \left (a+b \sin ^{-1}(c x)\right )^2}{\left (d-c^2 d x^2\right )^3} \, dx \]

Verification is Not applicable to the result.

[In]

Int[(x^m*(a + b*ArcSin[c*x])^2)/(d - c^2*d*x^2)^3,x]

[Out]

-(b*c*x^(2 + m)*(a + b*ArcSin[c*x]))/(6*d^3*(1 - c^2*x^2)^(3/2)) - (b*c*(1 - m)*x^(2 + m)*(a + b*ArcSin[c*x]))
/(6*d^3*Sqrt[1 - c^2*x^2]) - (b*c*(3 - m)*x^(2 + m)*(a + b*ArcSin[c*x]))/(4*d^3*Sqrt[1 - c^2*x^2]) + (x^(1 + m
)*(a + b*ArcSin[c*x])^2)/(4*d^3*(1 - c^2*x^2)^2) + ((3 - m)*x^(1 + m)*(a + b*ArcSin[c*x])^2)/(8*d^3*(1 - c^2*x
^2)) + (b*c*(1 - m)*(1 + m)*x^(2 + m)*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, c^2*x^2
])/(6*d^3*(2 + m)) + (b*c*(3 - m)*(1 + m)*x^(2 + m)*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, (2 + m)/2, (4 +
m)/2, c^2*x^2])/(4*d^3*(2 + m)) + (b^2*c^2*(1 - m)*x^(3 + m)*Hypergeometric2F1[1, (3 + m)/2, (5 + m)/2, c^2*x
^2])/(6*d^3*(3 + m)) + (b^2*c^2*(3 - m)*x^(3 + m)*Hypergeometric2F1[1, (3 + m)/2, (5 + m)/2, c^2*x^2])/(4*d^3*
(3 + m)) + (b^2*c^2*x^(3 + m)*Hypergeometric2F1[2, (3 + m)/2, (5 + m)/2, c^2*x^2])/(6*d^3*(3 + m)) - (b^2*c^2*
(1 - m)*(1 + m)*x^(3 + m)*HypergeometricPFQ[{1, 3/2 + m/2, 3/2 + m/2}, {2 + m/2, 5/2 + m/2}, c^2*x^2])/(6*d^3*
(6 + 5*m + m^2)) - (b^2*c^2*(3 - m)*(1 + m)*x^(3 + m)*HypergeometricPFQ[{1, 3/2 + m/2, 3/2 + m/2}, {2 + m/2, 5
/2 + m/2}, c^2*x^2])/(4*d^3*(6 + 5*m + m^2)) + ((1 - m)*(3 - m)*Defer[Int][(x^m*(a + b*ArcSin[c*x])^2)/(d - c^
2*d*x^2), x])/(8*d^2)

Rubi steps

\begin{align*} \int \frac{x^m \left (a+b \sin ^{-1}(c x)\right )^2}{\left (d-c^2 d x^2\right )^3} \, dx &=\frac{x^{1+m} \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac{(b c) \int \frac{x^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{\left (1-c^2 x^2\right )^{5/2}} \, dx}{2 d^3}+\frac{(3-m) \int \frac{x^m \left (a+b \sin ^{-1}(c x)\right )^2}{\left (d-c^2 d x^2\right )^2} \, dx}{4 d}\\ &=-\frac{b c x^{2+m} \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac{x^{1+m} \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{(3-m) x^{1+m} \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}+\frac{\left (b^2 c^2\right ) \int \frac{x^{2+m}}{\left (1-c^2 x^2\right )^2} \, dx}{6 d^3}-\frac{(b c (1-m)) \int \frac{x^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{6 d^3}-\frac{(b c (3-m)) \int \frac{x^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{4 d^3}+\frac{((1-m) (3-m)) \int \frac{x^m \left (a+b \sin ^{-1}(c x)\right )^2}{d-c^2 d x^2} \, dx}{8 d^2}\\ &=-\frac{b c x^{2+m} \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac{b c (1-m) x^{2+m} \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \sqrt{1-c^2 x^2}}-\frac{b c (3-m) x^{2+m} \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \sqrt{1-c^2 x^2}}+\frac{x^{1+m} \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{(3-m) x^{1+m} \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}+\frac{b^2 c^2 x^{3+m} \, _2F_1\left (2,\frac{3+m}{2};\frac{5+m}{2};c^2 x^2\right )}{6 d^3 (3+m)}+\frac{\left (b^2 c^2 (1-m)\right ) \int \frac{x^{2+m}}{1-c^2 x^2} \, dx}{6 d^3}+\frac{\left (b^2 c^2 (3-m)\right ) \int \frac{x^{2+m}}{1-c^2 x^2} \, dx}{4 d^3}+\frac{((1-m) (3-m)) \int \frac{x^m \left (a+b \sin ^{-1}(c x)\right )^2}{d-c^2 d x^2} \, dx}{8 d^2}+\frac{(b c (1-m) (1+m)) \int \frac{x^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{1-c^2 x^2}} \, dx}{6 d^3}+\frac{(b c (3-m) (1+m)) \int \frac{x^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{1-c^2 x^2}} \, dx}{4 d^3}\\ &=-\frac{b c x^{2+m} \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac{b c (1-m) x^{2+m} \left (a+b \sin ^{-1}(c x)\right )}{6 d^3 \sqrt{1-c^2 x^2}}-\frac{b c (3-m) x^{2+m} \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \sqrt{1-c^2 x^2}}+\frac{x^{1+m} \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{(3-m) x^{1+m} \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}+\frac{b c (1-m) (1+m) x^{2+m} \left (a+b \sin ^{-1}(c x)\right ) \, _2F_1\left (\frac{1}{2},\frac{2+m}{2};\frac{4+m}{2};c^2 x^2\right )}{6 d^3 (2+m)}+\frac{b c (3-m) (1+m) x^{2+m} \left (a+b \sin ^{-1}(c x)\right ) \, _2F_1\left (\frac{1}{2},\frac{2+m}{2};\frac{4+m}{2};c^2 x^2\right )}{4 d^3 (2+m)}+\frac{b^2 c^2 (1-m) x^{3+m} \, _2F_1\left (1,\frac{3+m}{2};\frac{5+m}{2};c^2 x^2\right )}{6 d^3 (3+m)}+\frac{b^2 c^2 (3-m) x^{3+m} \, _2F_1\left (1,\frac{3+m}{2};\frac{5+m}{2};c^2 x^2\right )}{4 d^3 (3+m)}+\frac{b^2 c^2 x^{3+m} \, _2F_1\left (2,\frac{3+m}{2};\frac{5+m}{2};c^2 x^2\right )}{6 d^3 (3+m)}-\frac{b^2 c^2 (1-m) (1+m) x^{3+m} \, _3F_2\left (1,\frac{3}{2}+\frac{m}{2},\frac{3}{2}+\frac{m}{2};2+\frac{m}{2},\frac{5}{2}+\frac{m}{2};c^2 x^2\right )}{6 d^3 \left (6+5 m+m^2\right )}-\frac{b^2 c^2 (3-m) (1+m) x^{3+m} \, _3F_2\left (1,\frac{3}{2}+\frac{m}{2},\frac{3}{2}+\frac{m}{2};2+\frac{m}{2},\frac{5}{2}+\frac{m}{2};c^2 x^2\right )}{4 d^3 \left (6+5 m+m^2\right )}+\frac{((1-m) (3-m)) \int \frac{x^m \left (a+b \sin ^{-1}(c x)\right )^2}{d-c^2 d x^2} \, dx}{8 d^2}\\ \end{align*}

Mathematica [A]  time = 9.12492, size = 0, normalized size = 0. \[ \int \frac{x^m \left (a+b \sin ^{-1}(c x)\right )^2}{\left (d-c^2 d x^2\right )^3} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(x^m*(a + b*ArcSin[c*x])^2)/(d - c^2*d*x^2)^3,x]

[Out]

Integrate[(x^m*(a + b*ArcSin[c*x])^2)/(d - c^2*d*x^2)^3, x]

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Maple [A]  time = 0.605, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{m} \left ( a+b\arcsin \left ( cx \right ) \right ) ^{2}}{ \left ( -{c}^{2}d{x}^{2}+d \right ) ^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(a+b*arcsin(c*x))^2/(-c^2*d*x^2+d)^3,x)

[Out]

int(x^m*(a+b*arcsin(c*x))^2/(-c^2*d*x^2+d)^3,x)

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Maxima [A]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{{\left (b \arcsin \left (c x\right ) + a\right )}^{2} x^{m}}{{\left (c^{2} d x^{2} - d\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(a+b*arcsin(c*x))^2/(-c^2*d*x^2+d)^3,x, algorithm="maxima")

[Out]

-integrate((b*arcsin(c*x) + a)^2*x^m/(c^2*d*x^2 - d)^3, x)

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Fricas [A]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}\right )} x^{m}}{c^{6} d^{3} x^{6} - 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} - d^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(a+b*arcsin(c*x))^2/(-c^2*d*x^2+d)^3,x, algorithm="fricas")

[Out]

integral(-(b^2*arcsin(c*x)^2 + 2*a*b*arcsin(c*x) + a^2)*x^m/(c^6*d^3*x^6 - 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 - d^3
), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*(a+b*asin(c*x))**2/(-c**2*d*x**2+d)**3,x)

[Out]

Timed out

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Giac [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{{\left (b \arcsin \left (c x\right ) + a\right )}^{2} x^{m}}{{\left (c^{2} d x^{2} - d\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(a+b*arcsin(c*x))^2/(-c^2*d*x^2+d)^3,x, algorithm="giac")

[Out]

integrate(-(b*arcsin(c*x) + a)^2*x^m/(c^2*d*x^2 - d)^3, x)